I read this question, and I would like to ask the opposite: Assume that I have a sequence of smoothing operators $(P_n)$ with (hence smooth) kernels $(p_n)$ converging strongly to some smoothing operator $P$ with kernel $p$. Does this imply convergence of $(p_n)$ to $p$, and if yes, in what sense?
Assume that I have $P_n \longrightarrow P$ strongly in $H^s$ for all $s \in \mathbb{R}$. Or conversely, which kind of convergence do I need for the operators to conclude that $p_n \longrightarrow p$ in $C^\infty$?
/Edit: As Liviu asked me to, I will be a bit more specific. It is ok for me to start with a compact manifold (everything is Riemannian), but I would like to have the results as well on a complete manifolds, with bounded curvature (or even with bounded geometry, if necessary, i.e. that all derivatives of the curvature tensor are uniformly bounded). It is enough to have the results on the kernel to be local though.
Regarding the convergence of my operator family, let us say that $\|P_n\|, \|P\| \leq C$ in $H^s_0(M)$ and that $P_nu \longrightarrow Pu$ for each $u \in C^\infty_c(M)$ (then we have the same statement for all $u \in H^s_0(M)$ be the Banach-Steinhaus Theorem).
Now the question is: If each $P_n$ is smoothing, as well as $P$, and the above holds for each $s \in \mathbb{R}$, can we conclude that the kernels converge $p_n$ converge to $p$ in $C^\infty(M \times M)$, at least locally?
